Vtyjkyuk

How to find (or estimate) $1.0003^365$ without using a calculator? Do we have any fast algorithm for cases where base is slightly more than one? Say up to $1.1$ with tick $0.05$.

You could try the binomial expansion of $(1+0.05)^10$ and stop calculating terms after they become small enough to not affect your required degree of accuracy

$$(1+x)^napprox 1+nxquad(xll1)$$

^{$$^*(1+x)^n=1+nx+fracn(n-1)2!x^2+fracn(n-1)(n-2)3!x^3+...$$}

Well binomial theorem with some basic algebraic manipulations,I find it easier to play with integers than decimals.
$$1.05^10=frac105^10100^10=frac5^10cdot21^105^10cdot20^10=frac(20+1)^1020^10=1+frac10cdot20^9+45cdot 20^8+10choose 320^7+cdots+120^10=1+1/2+frac4520^2+frac12020^3+cdots+frac120^10=1.5+0.1125+0.015+cdots+frac120^10approx 1.6275$$

First, learn how to work with base-10 logs in your head, including converting them back and forth. Here's a simple video that will teach you how to do this: https://www.youtube.com/watch?v=V5rgTPu8JcE

Next, realize that $e$ can be used to approximate such problems. A good understanding of $e$ can be found here: http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/

Finally, learn to take equations in the form of $e^x$ and convert them into equivalent problems of the form $10^y$: http://headinside.blogspot.com/2014/03/calculate-powers-of-e-in-your-head.html

To actually work through the problem, multiply the interest (0.05 in your example) times the power (10 in your example), and start by raising $e$ to that power. 0.05 × 10 = 0.50, so we're looking at roughly $e^0.50$.

We're going to convert this into the form $10^y$, but the conversion method works better with integers, so we should the equivalent of $e^50$, work out $y$ for $10^y$, and then move the decimal place of $y$ two spaces to the left to compensate.

Using the method from the calculate powers of e tutorial above, $e^50$≈$10^21.715$, so $e^0.50$≈$10^0.21715$. Now, we just have to work out what the base-10 antilogarithm for 0.21715.

As explained in the youtube link above, a 0.041 difference between logarithms corresponds to a roughly 10% change in the equivalent decimal. 0.217 + 0.041 + 0.041 = 0.299, which is quite close to the base-10 log of 2. Two differences of 0.041 mean two differences of 10%. In short, the answer should be roughly 20% less than 2, which is 1.6.

OK, this probably isn't a quick a method as you may like, but it does return a decent estimate.

You can use successive squaring.

$$x^10 = x^2(x^4)^2 = x^2((x^2)^2)^2$$

so first calculate $x^2$, then save it for later, calculate square two more times and then multiply with the $x^2$ you saved. This will make smaller the number of multiplications you need to do. From 9 down to 4. Will be much more impressive for large powers and powers that are a multiple of two. For example $x^1024$ will require no more than 10 multiplications!